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gregori [183]
3 years ago
13

At the same temperature, two wires made of pure copper have different resistances. The same voltage is applied at the ends of ea

ch wire. The wires may differ in Check all that apply. View Available Hint(s) Check all that apply.(A) length.(B) cross-sectional area.(C) resistivity.(D) amount of electric current passing through them.
Physics
2 answers:
vazorg [7]3 years ago
7 0

Answer:

A and B true, C and D false.

Explanation:

We only know that they have the same resistance and the same voltage is applied to them. We need to remember the formula of resistance of a wire:

R=\rho \frac{L}{A}

Where \rho is the resistivity (a property of the material), L its length and A its cross-sectional area. We could change L and A and still obtain the same R. From this we see that:

A) True. They could have any length.

B) True. They could have any cross-sectional area.

C) False. Resistivity is a property of the material, in this case, copper.

D) False. I=V/R, and both are the same for both wires.

Anastasy [175]3 years ago
3 0

Answer:

(A) length.(B) cross-sectional area.

Explanation:

The two options that they could have different are the length and the cross sectional area, and that can make the have the different resitances the longer the wire the more resitance it will have, since they are made from the same material and the same voltage is applied ot them, neither C or D could be because the resistivity depends solely on the material with which things are made of and both are pure copper wires.

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In an early attempt to understand atomic structure, Niels Bohr modeled the hydrogen atom as an electron in uniform circular moti
AleksandrR [38]

Answer:

The answer is below

Explanation:

Using Coulomb's law of electric field which is:

F=k\frac{q_1q_1}{r^2}\\\\ k =constant=9*10^9\ Nm^2/C^2,q_1=q_2=1.6*10^{-19}C,r=5.29*10^{-11}m\\\\Substituting\ gives:\\\\F=9*10^9*\frac{(1.6*10^{-19})*(1.6*10^{-19})}{(5.29*10^{-11})^2} =8.22*10^{-8}N\\\\Both\ centripetal\ force\ is\ given\ by:\\\\F=m\frac{v^2}{r} \\\\m = mass\ of \ electron=9.11*10^{-31}g,v=speed\ of\ electron\\\\F=m\frac{v^2}{r} \\\\v=\sqrt{\frac{F*r}{m} } \\\\subsituting:\\\\v=\sqrt{\frac{8.22*10^{-8}*5.29*10^{-11}}{9.11*10^{-31}} } \\\\v=2.18*10^6\ m/s\\\\

But\ \omega=\frac{v}{r}=\frac{2.18*10^6}{5.29*10^{-11}}  =4.13*10^{17}\\\\\omega=2\pi f; f=frequency\\\\f=\frac{\omega}{2\pi} =\frac{4.13*10^{17}}{2\pi} \\\\f=6.57*10^{15}\ Hz

6 0
3 years ago
CAN SOMEONE PLEASE HELPPPPPP
Natali5045456 [20]

Answer:

0.54

Explanation:

2.7÷5 = 0.54

Hope this helps!

6 0
3 years ago
4. O/N 15/P11/Q11
anygoal [31]

Answer:

B) 4500 Pa

Explanation:

As pressure is force per unit area,

P = F/A

It stands to reason that the smallest pressure for a given force is when it is shared by the largest area.

The possible areas are

0.30(0.40) = 0.12 m²

0.30(0.50) = 0.15 m²

0.40(0.50) = 0.20 m²

The pressure when the face with the largest area (0.20 m²) is down is

P = 900 / 0.20 = 4500 N/m² or 4500 Pa

the other possible pressures would be

900/0.15 = 6000 Pa

900/0.12 = 7500 Pa

which are both larger than our solution.

5 0
4 years ago
Jake has noticed his best friend Alison cheating at chess several times this week. If she gets caught, she will be kicked off th
tensa zangetsu [6.8K]
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8 0
4 years ago
Read 2 more answers
A uniform meter rule with a mass of 200g is suspended at zero mark pivotes at 22.0cm mark. calculate the mass of the rule.
denpristay [2]

Answer:

The mass of the rule is 56.41 g  

Explanation:

Given;

mass of the object suspended at zero mark, m₁ = 200 g

pivot of the uniform meter rule = 22 cm

Total length of meter rule = 100 cm

0                          22cm                                  100cm

-------------------------Δ------------------------------------

↓                                                                       ↓

200g                                                                 m₂  

Apply principle of moment

(200 g)(22 cm - 0)     = m₂(100 cm - 22 cm)

(200 g)(22 cm) = m₂(78 cm)

m₂ =  (200 g)(22 cm)  / (78 cm)

m₂ = 56.41 g  

Therefore,  the mass of the rule is 56.41 g                                            

3 0
4 years ago
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